Let $f$ be a non-negative, bounded and continuous function such that $\int_\mathbb{R} f(x)\, \mathrm{d}x < \infty$. Does it hold that $$ \lim_{n \rightarrow \infty} \sum_{j \in \mathbb{Z}} \frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) = \int_\mathbb{R} f(x)\, \mathrm{d}x $$
Current approach
Due to $f \geq 0$ we can use Fubini's theorem to show that $$ \sum_{j \in \mathbb{Z}} \frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) = \frac{1}{n} \sum_{k = 1}^n \sum_{j \in \mathbb{Z}} f\bigg(j + \frac{k}{n} \bigg). $$ Thus, the question boils down to whether the Riemann sum converges when the function that is being integrated is an infinite series, i.e. does it hold that $$ \frac{1}{n} \sum_{k = 1}^n \sum_{j \in \mathbb{Z}} f\bigg(j + \frac{k}{n} \bigg) \rightarrow \int_0^1 \sum_{j \in \mathbb{Z}} f(j + x) \, \mathrm{d}x $$ as $n \rightarrow \infty$?
My current approach to show this looks as follows: $$ \bigg\vert \frac{1}{n} \sum_{k = 1}^n \sum_{j \in \mathbb{Z}} f\bigg(j + \frac{k}{n} \bigg) - \int_0^1 \sum_{j \in \mathbb{Z}} f(j + x) \, \mathrm{d}x \bigg\vert \leq \sum_{j \in \mathbb{Z}} \bigg\vert \frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) - \int_0^1 f(j + x) \, \mathrm{d}x \bigg\vert $$ Here, I can't figure out how to bound the absolute value such that the limit of the series converges to zero.
Define $f$ (continuous nonnegative integrable) like this. Let $f(x) = 0$ on $(-\infty,1]$. For each $m \in \mathbb N$, on the interval $[m,m+1]$ the function has
$f\left(m+\frac{k}{m}\right) = 1$ for $k=1,2,\dots,m-1$,
$f(m)= f(m+1) = 0$,
$0\le f(x) \le 1$,
$\int_m^{m+1} f = \frac{1}{m^2}$.
For example, on $[4,5]$ it could look like this:
where the three triangles have height $1$ and width $1/24$.
Let $$ S_n = \sum_{j \in \mathbb{Z}} \frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) . $$ We claim $S_n$ does not converge to $\int_{-\infty}^\infty f$.
Choose $j_1 \in \mathbb N$ so that $$ \sum_{j=j_1}^\infty \frac{1}{j^2} < \frac14 \quad\text{and thus}\quad \int_{j_1}^\infty f < \frac14 . $$ Then choose $n_1 \ge 4$ so that, for each $j$ with $1 \le j \le j_1$, and for each $n \ge n_1$, $$ \left|\frac{1}{n}\sum_{k=1}^{n} f\left(j+\frac{k}{n}\right) - \int_{j}^{j+1} f\right| < \frac{1}{4j_1} \quad\text{and thus} \quad \sum_{j=1}^{j_1} \left|\frac{1}{n}\sum_{k=1}^{n} f\left(j+\frac{k}{n}\right) - \int_{j}^{j+1} f\right| < \frac{1}{4}. $$ We claim that, for all $n > n_1$, we have $$ \left|S_n - \int_{-\infty}^\infty f\right| > \frac{1}{4} $$ [and thus $S_n$ does not converge to $\int_{-\infty}^\infty f$]. To prove this: Fix $n > n_1$. Then $$ S_n = \sum_{j \in \mathbb{Z}} \frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) = A_n+B_n,\quad\text{where}\\ A_n=\sum_{j=1}^{j_1}\frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg)\\ B_n=\sum_{j=j_1+1}^{\infty}\frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) . $$
Compute: \begin{align} A_n - \int_1^{j_1+1} f &= \sum_{j=1}^{j_1}\left[\frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) - \int_j^{j+1} f\right] \\ & \ge -\sum_{j=1}^{j_1} \left|\frac{1}{n}\sum_{k=1}^{n} f\left(j+\frac{k}{n}\right) - \int_{j}^{j+1} f\right| > -\frac{1}{4} , \\ B_n& \ge \frac{1}{n} \sum_{k = 1}^n f\bigg(n + \frac{k}{n} \bigg) =\frac{n-1}{n} \ge \frac34 , \\ - \int_{j_1+1}^\infty f & \ge - \frac{1}{4}. \end{align} Add to get $$ S_n - \int_1^\infty f > \frac14 . $$